### Quantitative Aptitude Practice Questions (With Solutions) Set-27

Quantitative Aptitude Practice Questions (With Solutions) Set-27:
List of Practice Questions for Upcoming SSC CGL Tier II Exam was given here with solutions, candidates those who are preparing for those exams can use this material

1).If x=√3 - 1/√3 and y=√3 + 1/√3, then the value of (x/y) + (y2/x) is
a)    √3
b)    3√3
c)    16√3
d)    2√3

2).A number is first decreased by 20%. The decreased number is then, increased by 20%. The resulting number is less than the original number by 20. Then the original number is
a)    200
b)    400
c)    500
d)    600

3).If a+b=12, ab=22, then (a2+b2) is equal to
a)    188
b)    144
c)    34
d)    100

4).Length of a side of a square inscribed in a circle is a√2 units. The circumference of the circle is
a)    2 πa units
b)    πa units
c)    4 πa units
d)    2a /π units

5).The side of a rhombus are 10 cm are each and a diagonal measures 16 cm. Area of the rhombus is
a)    96 sq.cm
b)    160 sq.cm
c)    100 sq.cm
d)    40 sq.cm

6).The diameter of the front wheel of an engine is 2x cm and that of rear wheel is 2y cm. To cover the same distance, find the number of times the rear wheel will revolve when the front wheel revolves ‘n’ times.
a)    n/xy times
b)    yn/x times
c)    nx/y times
d)    xy/n times

7). The length, breath and height of a cuboid are in the ratio 3:4:6 and its volume is 576 cm3. The whole surface of the cuboid is
a)    216 cm2
b)    324 cm2
c)    432cm2
d)    460 cm2

8). The perimeter and length of a rectangle are 40 m and 12 m respectively. Its breath will be
a)    10m
b)    8m
c)    6m
d)    3m

9). If(x-2) is a factor of x2+3Qx-2Q, then the value of Q is
a)    2
b)    -2
c)    1
d)    -1

10). If PA and PB are two tangents to a circle with centre O such that ∠AOB =110, then ∠APB is
a)    90°
b)    70°
c)    60°
d)    55°

1). b) 2). c) 3). d) 4). a) 5). a) 6). c) 7). c) 8). b) 9). d) 10). b)

Check here the Solution for above Aptitude Questions:

1). x=√3-(1/√3) ; y=√3+(1/√3)
x+y =2√3
xy=[√3-(1/√3)] [√3+(1/√3)]
=3-1/3=8/3
x3+y3=(x+y)3-3xy (x+y)
= (2√3)3-2(8/3)(2√3)
=24√3-√3 =8√3
Now,
(x2/y) + (y2/x) =x3+y3/xy = 8√3 / (8/3) =3√3
2). Let the number be x.
Then 120% of (80% of x)= x-20
120/100 × 80/100x = x-20
→24x / 25 =x-20
24x=25x-500
x =500
3). a+b=12
ab  =22
(a+b)2 =a2+b2+2ab
122 =a2+b2+2×22
144 = a2+b2+44
a2+b2=144-44 =100
4).

DE =0E    =a√2 /2 =a√2
= √(a2/2 + a2/2) = √a2 =a
Circumference =2 πr =2 πα
5).

∆AOB is a right angled triangle.
Then AB2   =AO2+OB2
100 = AO2+64
AO2 =100 -64 =36
AO =6 CM
d1=BD =16
d2=AC=6+6 =12
Area =1/2(d1+d2
=1/2×16×12 =96cm2
6). Let the real wheel rotates N times
Then 2 πr ×n =2xy×N
→N=2πxn / 2πy =xn/y
7). Let the length, breath and height be 3x, 4x and 6x respectively
Volume = 3x×4x×6x =576
=72x3 =576
= x3 = 576/72
=x3=8
=x=2
Whole Surface Area = 2×(3x × 4x + 4x × 6x × 3x)
= 2( 12x2 + 24x2 + 18x2)
= 2 × 54x2
8). Length l=12m
Let breath be h m
Then Perimeter 2(l+h) =40
l+h =20
12+h =20-12 =8m
9). x-2  is a factor of x3+3Qx -2Q
→x=2 is a root of
x2+3Qx-2Q=0
→22+3Q(2)-2Q=0
4+6Q-2Q=0
4 Q = - 4
4Q=-4/4 =-1